Question

What is the unit of $k$ in the relation ${\text{U}} = \dfrac{{{\text{k}}y}}{{{y^2} + {a^2}}}$ where $U$ represents potential energy, $y$ represents the displacements and a represents the amplitude.

Answer 1

To solve this question one must be aware of dimensional equations. A dimensional equation is obtained when a physical quantity is equated with a dimensional formula. In this analysis base units such as mass, length and time are represented as$\left[ {\text{M}} \right]$,$\left[ {\text{L}} \right]$and $\left[ {\text{T}} \right]$respectively. Solve this question by determining the LHS and RHS in terms of dimensions and then by equating both the quantities.
Formula used: Dimensional analysis; Mass can be expressed as $\left[ {\text{M}} \right]$, Length can be expressed as $\left[ {\text{L}} \right]$ and Time can be expressed as $\left[ {\text{T}} \right]$

Complete step by step answer:
The equation that has been provided to us ${\text{U}} = \dfrac{{{\text{k}}y}}{{{y^2} + {a^2}}}$
Express the above equation in dimensional quantities
Mass can be expressed as $\left[ {\text{M}} \right]$
Length can be expressed as $\left[ {\text{L}} \right]$ and
Time can be expressed as $\left[ {\text{T}} \right]$
Here U is the potential energy. The unit of energy is Joules.
Now joules can be written as Newton-meter.
In mathematical terms;${\text{Joule}} = {\text{Newton meter}}$
$= {\text{mass}} \times {\text{acceleration}} \times {\text{meter}}$
Acceleration is meter per second squared. Hence, in dimensional terms we get$[{\text{L}}{{\text{T}}^{ - 2}}]$.
The equation${\text{U}} = \dfrac{{{\text{k}}y}}{{{y^2} + {a^2}}}$,
LHS in dimensional terms will be equal to $[{\text{M][L}}{{\text{T}}^{ - 2}}][{\text{L}}] = [{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$ …………….$(1)$
RHS in dimensional terms
$\dfrac{{{\text{k}}y}}{{{y^2} + {a^2}}}$, where, y-displacement and a-amplitude will be equal to $\dfrac{{{\text{k[L]}}}}{{{\text{[}}{{\text{L}}^{\text{2}}}{\text{] + [}}{{\text{L}}^{\text{2}}}]}}$ ……………… $(2)$
Both the LHS and RHS of an equation must be dimensionally equal to each other.
Therefore $(1)$and $(2)$ must be equal to each other
Therefore we get, $[{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}] = \dfrac{{{\text{k[L]}}}}{{{\text{[}}{{\text{L}}^{\text{2}}}{\text{] + [}}{{\text{L}}^{\text{2}}}]}}$
$[{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}] = \dfrac{{{\text{k[L]}}}}{{{\text{2[}}{{\text{L}}^{\text{2}}}]}}$
Here, to find the dimension of k we can ignore the constant$2$.
Therefore we get, $[{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}] = \dfrac{{{\text{k[L]}}}}{{{\text{[}}{{\text{L}}^{\text{2}}}]}}$
By simplifying and cross-multiplying we get, $\dfrac{{\text{k}}}{{{\text{[}}{{\text{L}}^{}}]}} = [{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$
${\text{k}} = [{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}][{\text{L}}]$
Here, $[{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$represents joules and $[{\text{L}}]$represents meter.
Therefore, the unit of $k$ is Joule-meter

Note: While mentioning the dimensions of physical quantities the constants in the equations can be ignored. Another imperative note is that only those physical quantities that have the same dimensions can be added together or subtracted from each other. Also using dimensions cannot be the only basis by which a formula can be deduced as right or wrong.